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Chapter 11: Trigonometric Identities

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11.1Trigonometric Identities

11.2Addition and Subtraction Formulas

11.3Double-Angle, Half-Angle, and Product-Sum Formulas

11.4Inverse Trigonometric Functions

11.5Trigonometric Equations

Derive the identity for cos(A – B).

Let angles A and B be angles in standard position on a unit circle with B < A.

Let Sand Q be the points on the terminal sides of angels A and B, respectively.

Q has coordinates (cosB, sin B).

S has coordinates (cosA, sin A).

R has coordinates (cos (A – B), sin (A – B)).

Angle SOQ equals A – B.

Since SOQ = POR, chords PR and SQ are equal.

- By the distance formula, chords PR = SQ,
Simplifying this equation and using the identity

sin² x + cos² x =1, we rewrote the equation as

cos(A– B) = cosAcosB + sin A sin B.

- To find cos(A + B), rewrite A + B as A– (–B) and use the identity for cos (A–B).

Cosine of a Sum Or Difference

cos(A – B) = cos A cos B + sin A sin B

cos(A + B) = cos A cos B– sin A sin B

ExampleFind the exact value of the following.

- cos 15°

Using the cofunction relationship and letting = A + B, we can obtain the:

Sine of a Sum or Difference

sin(A + B) = sin A cosB + cosA sin B

sin(A–B) = sin A cosB –cosA sin B

- Using the identities for sin(A+ B), cos(A+ B), and tan(–B) = –tan B, we can derive the identities for the tangent of a sum or difference.

Tangent of a Sum or Difference

ExampleFind the exact value of the following.

- sin 75°
- tan
- sin 40° cos 160° – cos 40° sin 160°
Solution

(a)

(b)

(c) sin 40° cos 160° – cos 40° sin 160°

= sin(40° – 160°)

= sin(–120°)

ExampleSuppose that A and B are angles in standard

position, with

Find each of the following.

- sin(A + B) (b) tan (A + B) (c) the quadrant of A + B
Solution

(a)

Since cosA < 0 in Quadrant II.

(b) Use the values of sine and cosine from part (a) to

get

(c) From the results of parts (a) and (b), we find that

sin(A + B) is positive and tan(A + B) is also positive. Therefore, A + B must be in quadrant I.

ExampleCommon household electric current is called

alternating current because the current alternates direction

within the wire. The voltage V in a typical 115-volt outlet can be

expressed by the equation V = 163 sin t, where is the

angular velocity (in radians per second) of the rotating generator

at the electrical plant and t is time measured in seconds.

- It is essential for electric generators to rotate at 60 cycles per second so household appliances and computers will function properly. Determine for these electric generators.
- Graph V on the interval 0 t .05.
- For what value of will the graph of V = 163cos(t–) be the same as the graph of V = 163 sin t?

Solution

- Since each cycle is 2 radians, at 60 cycles per second,
= 60(2) = 120 radians per second.

(b)V = 163 sin t = 163 sin 120t.

Because amplitude is 163,

choose –200 V 200 for the

range.

(c)